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JEE Main & Advanced Sample Paper JEE Main Sample Paper-7

  • question_answer
    A charged particle enters a unifrom magnetic field at an `angle of \[45{}^\circ \] with the magnetic field. The pitch of the helical path followed is P. The radius of the helix will be

    A)  \[\frac{P}{\sqrt{2\pi }}\]                             

    B)  \[\sqrt{2P}\]

    C)  \[\frac{P}{2\pi }\]                                           

    D)  \[\frac{\sqrt{2}P}{\pi }\]

    Correct Answer: C

    Solution :

    Pitch \[=\upsilon \,\cos \theta \times \frac{2\pi r}{\upsilon }=2\pi r\,\cos \theta =2\pi r=\frac{1}{\sqrt{2}}\] \[\therefore \] Radius of circle \[r=\frac{P\sqrt{2}}{2\pi }=\frac{P}{\sqrt{2}\pi }\] Radius of helix \[=r\cos {{45}^{o}}=\frac{P}{2\pi }\]


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